3

在 T-SQL 中，我只需在 select 语句中使用 group by 子句和 count(*) 来给我需要的值。但是对于多维数据集，情况就不同了，因为计数不仅仅是行数，而是维度组合。所以我用谷歌搜索了一个无济于事的答案。这是我的问题的详细解释：

我原来的MDX是：

SELECT 
  NON EMPTY 
    {
        [Measures].[Budget]
    } ON COLUMNS
 ,NON EMPTY 
    {
        [Location].[Category - Entity - Facility].[Facility].ALLMEMBERS*
        [Location].[Category - Facility - Unit].[Location].ALLMEMBERS*
        [Calendar].[Day].[Day].ALLMEMBERS
    } ON ROWS
FROM 
(
  SELECT 
    {[Location].[Category - Entity - Facility].[Category].&[3]} ON COLUMNS
  FROM 
  (
    SELECT 
        [Calendar].[Year  -  Quarter -  Month -  Day].[Day].&[2012-01-01T00:00:00]
      : [Calendar].[Year  -  Quarter -  Month -  Day].[Day].&[2012-05-31T00:00:00]
    ON COLUMNS
    FROM [PHI Census]
  )
)

Results look like this:
    Facility 1    Location 1    Day 1    100
    Facility 1    Location 1    Day 2    100
    Facility 1    Location 1    Day 3    100
    Facility 1    Location 1    Day 4    100
    Facility 1    Location 2    Day 1    80
    Facility 1    Location 2    Day 2    80
    Facility 1    Location 2    Day 3    80
    Facility 2    Location 1    Day 1    65
    Facility 2    Location 1    Day 2    65
    Facility 2    Location 1    Day 3    65
    Facility 2    Location 1    Day 4    65
    Facility 2    Location 2    Day 1    73
    Facility 2    Location 2    Day 2    73
    Facility 2    Location 2    Day 3    73

这给了我为每个设施-位置-日组合列出一次的 [预算]。我想从 ON ROWS 子句中删除 [Calendar].[Day].[Day].ALLMEMBERS 并简单地使用一个计算成员，该成员将返回每个 Facility-Location 组合以及每一行的天数。所以基本上，

The results would look like this:
    Facility      Location      Budget   DayCount
    Facility 1    Location 1    100      4
    Facility 1    Location 2    80       3
    Facility 2    Location 1    65       4
    Facility 2    Location 2    73       3
4

1 回答 1

3

DayCount 的表达式可以是：

MEMBER [Measures].[DayCount] AS Count(NonEmpty([Calendar].[Day].[Day].ALLMEMBERS, [Measures].[Budget]))
于 2012-07-12T15:32:11.577 回答